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yaroslaw [1]
3 years ago
8

Passengers in a high-flying jumbo jet feel their normal weight in flight, while passengers in an orbiting space vehicle do not,

because passengers in the space vehicle are
A) Without support forces.
B) Beyond the main pull of Earth's gravity
C) above Earth's atmosphere
D) all of the above
E) None of the above.
Physics
1 answer:
daser333 [38]3 years ago
4 0

Answer:E none of the above

Explanation:

When a space shuttle orbiting around the earth then earth gravitational force keeps the shuttle in its orbit by providing centripetal force. Tangential velocity makes the shuttle to revolve around the earth. Earth gravitational pull is significant to make the shuttle move in orbit while the Passengers feel weightlessness because they are falling downward but never able to reach as they move fast enough such that earth curves away them.                        

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What is keeping all variables the same except for the one being tested an example of?
stepan [7]

Answer:

A fair test.

Explanation:

Hi, a fair test is used to do scientifically valuable experiments, is a controlled investigation to answer a scientific question.

In a fair test two or more things are compared.

It consists in changing only one factor (the one bieng tested) and keeping all the other conditions the same during an experiment.

The factor is called a variable.

8 0
3 years ago
Read 2 more answers
A psychopathologist records the number of criminal offenses among teenage drug users in a nationwide sample of 1,201 participant
lara [203]

Answer:

Standard deviation = 3

Explanation:

Given

N = 1201

SS = 10800

Required

Determine the standard deviation

First, we need to determine the variance;

Variance = \frac{SS}{N - 1}

This gives:

Variance = \frac{10800}{1201 - 1}

Variance = \frac{10800}{1200}

Variance = 9

Know that:

Variance = SD^2

Where SD represents standard deviation

This gives

9 = SD^2

Take square root

SD = \sqrt 9

SD = 3

7 0
3 years ago
"why are some galaxies' spectra blueshifted rather than redshifted?"
stira [4]
This phenomena is also called the Doppler shift. When the source of light is approaching towards an observer, the color tends to be blue shifted, but when the source is moving away or being stretch, the color tends to red shifted. In astronomy it can be use how fast galaxy is moving towards us or how fast it moves away.
8 0
3 years ago
Light with a wavelength of 400 nm strikes the surface of cesium in a photocell, and the maximum kinetic energy of the electrons
Firdavs [7]

Answer:

The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

Explanation:

Given that,

Wavelength = 400 nm

Energy E=1.54\times10^{-19}\ J

We need to calculate the longest wavelength of light that is capable of ejecting electrons from that metal

Using formula of energy

E = \dfrac{hc}{\lambda}

\lambda=\dfrac{hc}{E}

Put the value into the formula

\lambda=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.54\times10^{-19}}

\lambda=1292\times10^{-9}\ m

\lambda=1292\ nm

Hence, The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

8 0
3 years ago
Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.
olganol [36]

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

U_{g,1} + K_{x,1} + K_{y,1} =  U_{g,2} + K_{x,2} + K_{y,2}

Where:

U_{g,1}, U_{g,2} - Initial and final gravitational potential energy, measured in joules.

K_{x,1}, K_{x,2} - Initial and final horizontal translational kinetic energy, measured in joules.

K_{y,1}, K_{y,2} - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0

g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})

y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}

Where:

y_{1}. y_{2} - Initial and final height of the arrow, measured in meters.

v_{y,1}, v_{y,2} - Initial and final vertical speed of the arrow, measured in meters.

g - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

v_{y,1} = v_{1}\cdot \sin \theta

Where:

v_{1} - Magnitude of the initial velocity, measured in meters per second.

\theta - Initial angle, measured in sexagesimal degrees.

If v_{1} = 82\,\frac{m}{s} and \theta = 24^{\circ}, the initial vertical speed is:

v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}

v_{y,1} \approx 33.352\,\frac{m}{s}

If g = 9.807\,\frac{m}{s^{2}}, v_{y,1} \approx 33.352\,\frac{m}{s} and v_{y,2} = 0\,\frac{m}{s}, the maximum height of the first arrow is:

y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }

y_{2} - y_{1} = 56.712\,m

Second arrow

U_{g,1} + K_{y,1} =  U_{g,3} + K_{y,3}

Where:

U_{g,1}, U_{g,3} - Initial and final gravitational potential energy, measured in joules.

K_{y,1}, K_{y,3} - Initial and final vertical translational kinetic energy, measured in joules.

m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0

g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})

y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}

If g = 9.807\,\frac{m}{s^{2}}, v_{y,1} = 82\,\frac{m}{s} and v_{y,3} = 0\,\frac{m}{s}, the maximum height of the first arrow is:

y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }

y_{3} - y_{1} = 342.816\,m

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

E = U + K_{x}

The expression is now expanded:

E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}

Where v_{x} is the horizontal speed of the arrow, measured in meters per second.

v_{x} = v_{1}\cdot \cos \theta

If v_{1} = 82\,\frac{m}{s} and \theta = 24^{\circ}, the horizontal speed is:

v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}

v_{x} \approx 74.911\,\frac{m}{s}

If m = 0.06\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{max} = 56.712\,m and v_{x} \approx 74.911\,\frac{m}{s}, the total mechanical energy is:

E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}

E = 201.720\,J

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

E = m\cdot g \cdot y_{max}

m = 0.06\,kg, g = 9.807\,\frac{m}{s^{2}} and y_{max} = 342.816\,m

E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)

E = 201.720\,J

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

7 0
3 years ago
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